A completely inelastic collision occurs between two balls ofwet putty that move

Question

A completely inelastic collision occurs between two balls ofwet putty that move directly toward each other along a verticalaxis. Just before the collision, one ball of mass 3.0 kg is movingupward at 20 m/s, and the other ball, of mass 1.4 kg, is moving downward at 15 m/s. How high do the combined two balls of puttyrise above the collision point? (Neglect air drag.)
i am able to find pf am unable to find theheight? A completely inelastic collision occurs between two balls ofwet putty that move directly toward each other along a verticalaxis. Just before the collision, one ball of mass 3.0 kg is movingupward at 20 m/s, and the other ball, of mass 1.4 kg, is moving downward at 15 m/s. How high do the combined two balls of puttyrise above the collision point? (Neglect air drag.)
i am able to find pf am unable to find theheight?

Explanation / Answer

In an inelastic collision :                v   = ( m 1 u1 + m2 u2 ) / ( m1 + m2 )    As the masses moves in oppositedirection after collision ,                v  = ( 3.0 * 20  -   1.4 * 15 ) / ( 3.0 + 1.4)          since downward direction is taken asnegative                                   =   8.863 m /s   up (b)           h = v2 / 2g              = ( 8.863 )2 / ( 2 * 9.8 )              = 4.00 m Hope this helps u!

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